That \(L^2\)  is a graph-like \(\varphi \) -set follows from the fact that \(\lbrace \varphi (x;a_i,b_i):i\in \lbrace 0,1,\dots ,n\rbrace \rbrace \)  is consistent if and only if the interval \((\max _i a_i,\min _j b_j)\)  is non-empty and that this is equivalent to the intersection \((a_i,b_i)\cap (a_j,b_j)\)  being nonempty for all \(i,j\in \lbrace 0,1,\dots ,n\rbrace \) .
The statement of the lemma is now equivalent to asking that the above graphs are the \(\le \) -minimal graphs that cannot arise in the following way: the vertices are a finite collection of possibly repeated non-empty open intervals in \(\mathbb {L}\)  and there is an edge between two intervals if and only if their intersection is non-empty.
This result (with distinct intervals) is proved in the particular case that \(\mathbb {L}=\mathbb {R}\)  by Lekkekerker and Boland [1]}. Our strategy will be to reduce the general case to the case of distinct intervals in \(\mathbb {R}\) . That is, we show that \(\mathrm {isoSh}(\mathbb {L}^2)\)  is equal to the set of isomorphism types of interval intersection graphs in \(\mathbb {R}\)  arising from distinct intervals.
First, we note that we may assume that \(\mathbb {L}\)  contains \((\omega ,<)\)  as a suborder (if not, the reverse order on \(\mathbb {L}\)  works). Suppose that \(\mathbb {G}\)  is the intersection graph for the collection of open intervals \(\lbrace (a_i,b_i):i\in \lbrace 0,1,\dots ,n\rbrace \rbrace \)  where \(a_i,b_i\in \mathbb {R}\) . We may then construct a set of intervals in \(\omega \)  by stretching out all of the \((a_i,b_i)\)  and translating them to the positive half of \(\mathbb {R}\)  so that the images of \((a_i,b_i)\)  and \((a_j,b_j)\)  intersect in \(\omega \)  if and only if \((a_i,b_i)\cap (a_j,b_j)\ne \emptyset \)  and hence our new intervals have an intersection graph isomorphic to the intersection graph of the \((a_i,b_i)\) . Mapping these intervals to intervals of \(\mathbb {L}\)  via the inclusion \(\omega \hookrightarrow \mathbb {L}\)  does not change the intersection graph so long as we did not allow intersections to look like \((n,n+1)\)  in \(\omega \)  (this can be achieved by choosing how much to stretch out the original intervals in \(\mathbb {R}\) ). Thus the set of intersection graphs in \(\mathbb {L}\)  contains all of the intersection graphs in \(\mathbb {R}\) .
For the other inclusion, suppose that \(\mathbb {G}=(\lbrace I_i:i\in \lbrace 0,1,\dots ,n\rbrace \rbrace ,E)\)  is the intersection graph for the (possible repeating) non-empty intervals \(I_i\)  in \(\mathbb {L}\) . Let \(I_i=(a_i,b_i)\)  and, for each \(\lbrace I_i,I_j\rbrace \in E\) , choose an element \(e_{ij}\in I_i\cap I_j\) . Define an order \(\mathbb {M}\)  with underlying set the union of the sets
\(&\lbrace a_i:i\in \lbrace 0,1,\dots ,n\rbrace \rbrace \\&\lbrace b_i:i\in \lbrace 0,1,\dots ,n\rbrace \rbrace \\&\lbrace e_{ij}:i<j,\lbrace I_i,I_j\rbrace \in E\rbrace \) 

and with ordering given by \(<^\mathbb {L}\) . \(\mathbb {M}\)  is finite and so isomorphic to an initial segment of \(\omega \) . The map \(\mathbb {M}\hookrightarrow \omega \hookrightarrow \mathbb {R}\)  sends the intervals \(I_i\)  to intervals of \(\mathbb {R}\)  having the same intersection graph as the \(I_i\) . If the intervals \(I_{i_0},\dots ,I_{i_m}\)  are equal (and none of the other intervals are equal to \(I_{i_0}\) ), then we may take the distinct intervals
\(\left\lbrace \left(a_{i_0}+\frac{j}{m+1},b_{i_0}-\frac{j}{m+1}\right):j\in \lbrace 0,\dots ,m\rbrace \right\rbrace \) 
to be the intervals corresponding to the \(I_{i_j}\)  without changing the intersection graph.
